What will be the ratio of the equivalent electromotive force (emf) of two cells connected in series to the equivalent emf of two cells connected in parallel?

A circuit consists of three batteries of emf $E_{1} = 1 \text{ V}$, $E_{2} = 2 \text{ V}$, and $E_{3} = 3 \text{ V}$ and internal resistances $1 \Omega$, $2 \Omega$, and $1 \Omega$ respectively, which are connected in parallel as shown in the figure. The potential difference between points $P$ and $Q$ is: (in $\text{ V}$)

In the previous problem,if the cells had been connected in parallel (instead of in series),which of the above graphs would have shown the relationship between total current $I$ and $n$ (number of cells)?

$n$ identical cells,each of emf $E$ and internal resistance $r$,are joined in series to form a row. $m$ such rows are joined in parallel across a load resistance $R$. The current in each cell is:

Two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition (such that $E_{1} > E_{2}$). If $r_{1}$ and $r_{2}$ are the internal resistances and $R$ is the external resistance,then the terminal potential difference across the external resistance $R$ is:

The potential difference in open circuit for a cell is $2.2\, V$. When a $4\, \Omega$ resistor is connected between its two electrodes,the potential difference becomes $2\, V$. The internal resistance of the cell will be .............. $\Omega$.